All practice questions

A sample containing

8.0\text{ g}

of impure calcium trioxocarbonate(IV) reacts completely with excess dilute hydrochloric acid according to the equation:

\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

If

1.344\text{ dm}^3

of carbon(IV) oxide gas measured at s.t.p. is liberated, what is the percentage purity of the calcium trioxocarbonate(IV) sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

75.0\%

70.0\%

25.0\%

60.0\%

Show answer & explanation

Answer: $75.0\%$

Answer

The percentage purity of the calcium trioxocarbonate(IV) sample is

75.0\%

.

First, find the moles of carbon(IV) oxide produced at s.t.p.:

1.344 / 22.4 = 0.060\text{ mol}

. According to the balanced chemical equation,

1\text{ mol}

of calcium trioxocarbonate(IV) reacts to yield

1\text{ mol}

of carbon(IV) oxide. Therefore,

0.060\text{ mol}

of pure calcium trioxocarbonate(IV) reacted. The mass of pure calcium trioxocarbonate(IV) is

0.060 \times 100\text{ g mol}^{-1} = 6.0\text{ g}

. Calculating percentage purity gives

(6.0\text{ g} / 8.0\text{ g}) \times 100\% = 75.0\%

.

Step-by-Step Solution

1

Calculate the moles of carbon(IV) oxide gas produced at s.t.p.

\text{Moles of CO}_2 = \frac{1.344\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.060\text{ mol}

Molar gas volume at s.t.p. equals

22.4\text{ dm}^3\text{ mol}^{-1}

.

2

Determine the mole relation and calculate the mass of pure calcium trioxocarbonate(IV).

Molar mass of

\text{CaCO}_3 = 40 + 12 + (3 \times 16) = 100\text{ g mol}^{-1}

. From the

1:1

stoichiometric ratio,

\text{moles of pure CaCO}_3 = 0.060\text{ mol}

.

\text{Mass} = 0.060\text{ mol} \times 100\text{ g mol}^{-1} = 6.0\text{ g}

.

The balanced chemical equation shows a 1:1 mole ratio between calcium trioxocarbonate(IV) and carbon(IV) oxide.

3

Calculate the percentage purity of the sample.

\text{Percentage purity} = \frac{\text{Mass of pure CaCO}_3}{\text{Total mass of impure sample}} \times 100\% = \frac{6.0\text{ g}}{8.0\text{ g}} \times 100\% = 75.0\%

Percentage purity expresses the mass fraction of pure active component relative to total sample mass.

Key Concept

Mass-volume stoichiometric calculations and percentage purity determination

Estimated Time:2m 0s

Question 4482Question →

A composite furnace wall consists of two tightly joined layers of equal cross-sectional area. Layer 1 has a thickness of $0.02\text{ m}$ and thermal conductivity of $200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . Layer 2 has a thickness of $0.04\text{ m}$ and thermal conductivity of $100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . The outer surface of Layer 1 is maintained at $120^\circ\text{C}$ and the outer surface of Layer 2 is maintained at $20^\circ\text{C}$ . Under steady-state conditions, what is the rate of heat transfer per unit area through the composite wall?

200\text{ kW}\cdot\text{m}^{-2}

250\text{ kW}\cdot\text{m}^{-2}

500\text{ kW}\cdot\text{m}^{-2}

100\text{ kW}\cdot\text{m}^{-2}

Show answer & explanation

Answer: $200\text{ kW}\cdot\text{m}^{-2}$

Answer

The rate of heat transfer per unit area through the composite furnace wall is

200\text{ kW}\cdot\text{m}^{-2}

.

Under steady-state conduction through composite layers in series, the total thermal resistance per unit area is the sum of the individual thermal resistances:

R_{\text{total}} = \frac{d_1}{k_1} + \frac{d_2}{k_2} = 1.0 \times 10^{-4} + 4.0 \times 10^{-4} = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

. Applying Fourier's law of heat conduction, the heat flux is

\frac{\Delta T}{R_{\text{total}}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}

.

Step-by-Step Solution

1

Calculate the thermal resistance per unit area (

R

) for each layer using

R_i = \frac{d_i}{k_i}

.

R_1 = \frac{0.02\text{ m}}{200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 1.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

and

R_2 = \frac{0.04\text{ m}}{100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 4.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

.

Thermal resistance quantifies a material's opposition to conductive heat flow per unit cross-sectional area.

2

Find the total thermal resistance per unit area (

R_{\text{total}}

) by summing the series resistances.

R_{\text{total}} = R_1 + R_2 = (1.0 \times 10^{-4}) + (4.0 \times 10^{-4}) = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

.

Layers in series add their thermal resistances directly.

3

Determine the rate of heat flow per unit area

\frac{Q/t}{A} = \frac{\Delta T}{R_{\text{total}}}

.

\frac{Q/t}{A} = \frac{120^\circ\text{C} - 20^\circ\text{C}}{5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}

.

At steady state, the heat flux across the composite structure is driven by the overall temperature gradient divided by total thermal resistance.

Key Concept

Thermal Resistance in Composite Layers (Conduction)

Estimated Time:2m 0s

Question 4483Question →

A series alternating current (AC) circuit contains an inductor of inductance $L = \frac{1}{\pi^2}\ \text{H}$ , a capacitor of capacitance $C = 25\ \mu\text{F}$ , and a resistor of resistance $R = 50\ \Omega$ . What is the resonant frequency of the circuit in hertz ( $\text{Hz}$ )?

Show answer & explanation

Answer: 100

Answer

The resonant frequency of the circuit is

100\ \text{Hz}

.

At resonance, the inductive reactance

X_L = 2\pi f L

equals the capacitive reactance

X_C = \frac{1}{2\pi f C}

. Equating both yields

f_0 = \frac{1}{2\pi\sqrt{LC}}

. Substituting

L = \frac{1}{\pi^2}\ \text{H}

and

C = 25 \times 10^{-6}\ \text{F}

gives

\sqrt{LC} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

, leading to

f_0 = \frac{1}{2\pi \left(\frac{5 \times 10^{-3}}{\pi}\right)} = 100\ \text{Hz}

.

Step-by-Step Solution

1

Write down the formula for the resonant frequency of a series RLC circuit.

f_0 = \frac{1}{2\pi\sqrt{LC}}

Resonance occurs when the inductive reactance equals the capacitive reactance (

X_L = X_C

).

2

Substitute the values of inductance

L = \frac{1}{\pi^2}\ \text{H}

and capacitance

C = 25 \times 10^{-6}\ \text{F}

into

\sqrt{LC}

.

\sqrt{LC} = \sqrt{\frac{1}{\pi^2} \times 25 \times 10^{-6}} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

Simplifying the square root removes the fraction containing

\pi

.

3

Calculate the resonant frequency

f_0

.

f_0 = \frac{1}{2\pi \times \frac{5 \times 10^{-3}}{\pi}} = \frac{1}{10^{-2}} = 100\ \text{Hz}

The factor of

\pi

cancels out in the denominator, resulting in a whole number value.

Key Concept

Resonant Frequency in AC Series Circuits

Estimated Time:1m 30s

Question 4484Question →

Given the function $y = \frac{x^2 + 2x}{(3x - 1)^2}$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

-\frac{5}{4}

\frac{1}{4}

\frac{13}{4}

\frac{1}{3}

Show answer & explanation

Answer: $-\frac{5}{4}$

Answer

-\frac{5}{4}

Applying the quotient rule

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

with

u = x^2 + 2x

(

u' = 2x + 2

) and

v = (3x - 1)^2

(

v' = 6(3x - 1)

) yields

\frac{dy}{dx} = \frac{16 - 36}{16} = -\frac{5}{4}

at

x = 1

.

Step-by-Step Solution

1

Identify the numerator and denominator functions for the quotient rule

y = \frac{u}{v}

.

u = x^2 + 2x

and

v = (3x - 1)^2

.

The function is structured as a quotient of two algebraic expressions.

2

Differentiate

u

and

v

with respect to

x

.

\frac{du}{dx} = 2x + 2

and

\frac{dv}{dx} = 2(3x - 1) \cdot 3 = 6(3x - 1)

.

Use the power rule for

u

and the chain rule for

v

.

3

Apply the quotient rule formula

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

.

\frac{dy}{dx} = \frac{(3x - 1)^2 (2x + 2) - (x^2 + 2x) \cdot 6(3x - 1)}{(3x - 1)^4}

.

The quotient rule formula combines the expressions and their derivatives.

4

Substitute

x = 1

into the derivative expression and simplify.

\frac{dy}{dx} = \frac{(2)^2 (4) - (3) \cdot 6(2)}{(2)^4} = \frac{16 - 36}{16} = -\frac{20}{16} = -\frac{5}{4}

.

Evaluating at

x = 1

yields the numerical derivative value.

Key Concept

Quotient Rule and Chain Rule of Differentiation

Question 4485Question →

Which of the following open-chain isomeric alkenes with the molecular formula $C_5H_{10}$ exhibits geometric (cis-trans) isomerism?

Pent-2-ene

Pent-1-ene

2-Methylbut-2-ene

3-Methylbut-1-ene

Show answer & explanation

Answer: Pent-2-ene

Answer

Pent-2-ene is the correct answer because neither of its double-bonded carbon atoms carries two identical attached groups.

Geometric (cis-trans) isomerism requires restricted rotation around a carbon-carbon double bond (

C=C

) along with two different substituents attached to each of the double-bonded carbon atoms. In pent-2-ene (

CH_3-CH=CH-CH_2-CH_3

), C2 is bonded to

-H

and

-CH_3

, while C3 is bonded to

-H

and

-CH_2CH_3

. Because neither carbon carries identical groups, pent-2-ene can form distinct cis and trans stereoisomers.

Step-by-Step Solution

1

Identify the structural condition required for geometric (cis-trans) isomerism in alkenes.

For an alkene

R_1R_2C=CR_3R_4

to exhibit geometric isomerism,

R_1 \neq R_2

and

R_3 \neq R_4

must hold for both carbon atoms involved in the double bond.

If either carbon atom in the double bond is bonded to two identical atoms or groups, rotating structural representation does not create distinct non-superimposable stereoisomers.

2

Analyze the structural formula of Pent-1-ene.

Pent-1-ene is

CH_2=CH-CH_2-CH_2-CH_3

. C1 is attached to two hydrogen atoms (

-H

and

-H

).

Since C1 has identical hydrogen atoms, it cannot exhibit cis-trans isomerism.

3

Analyze the structural formula of Pent-2-ene.

Pent-2-ene is

CH_3-CH=CH-CH_2-CH_3

. C2 is bonded to

-H

and

-CH_3

. C3 is bonded to

-H

and

-CH_2CH_3

.

Both double-bonded carbons have two distinct attached groups, permitting the existence of cis-pent-2-ene and trans-pent-2-ene.

4

Analyze the structural formulas of 2-Methylbut-2-ene and 3-Methylbut-1-ene.

2-Methylbut-2-ene has two methyl groups on C2 (

(CH_3)_2C=CH-CH_3

), and 3-Methylbut-1-ene has two hydrogen atoms on C1 (

CH_2=CH-CH(CH_3)_2

).

Both compounds violate the condition of having distinct groups on each doubly bonded carbon atom.

Key Concept

Geometric (cis-trans) Isomerism in Alkenes

Estimated Time:1m 30s

Question 4486Question →

During photosynthesis in vascular plants, synthesized organic nutrients such as sucrose are distributed from the leaves to roots, fruits, and growing points. Which plant tissue is primarily responsible for carrying out this transport process?

Phloem

Xylem

Cortex

Sclerenchyma

Show answer & explanation

Answer: Phloem

Answer

Phloem

Phloem is the specialized vascular tissue responsible for translocation, moving sugars and organic compounds manufactured in leaves to non-photosynthetic organs such as roots, stems, and fruits.

Step-by-Step Solution

1

Identify the type of substance being transported in the plant scenario

The scenario describes the transport of manufactured organic nutrients (sucrose) from photosynthetic leaves to other plant organs.

Distinguishing between organic solutes and inorganic raw materials determines the specific vascular tissue responsible.

2

Match the transport material with the appropriate vascular tissue

Phloem tissue (comprising sieve tube elements and companion cells) is specialized for translocating organic solutes from source to sink.

Xylem transports water and dissolved minerals, while phloem translocates photosynthates.

Key Concept

Function of Vascular Tissues in Plant Transport Systems

Question 4487Question →

An ecologist studying a freshwater habitat needs to determine the depth of light penetration (turbidity) in the water body and the relative humidity of the surrounding air. Which pair of instruments should be selected for these respective measurements?

Secchi disc and hygrometer

Hydrometer and wind vane

Anemometer and Secchi disc

Barometer and hydrometer

Show answer & explanation

Answer: Secchi disc and hygrometer

Answer

Secchi disc and hygrometer

The Secchi disc is a circular disc used in aquatic ecology to gauge water clarity and the depth to which solar radiation penetrates. Relative humidity, which measures moisture in the atmosphere, is recorded using a hygrometer.

Step-by-Step Solution

1

Identify the instrument used to measure water transparency and light penetration depth in aquatic habitats.

A Secchi disc is lowered into the water until it is no longer visible to measure turbidity/light penetration depth.

Light penetration in water bodies directly influences aquatic plant photosynthesis and distribution.

2

Identify the instrument used to measure atmospheric humidity.

A hygrometer (or wet-and-dry bulb psychrometer) measures relative humidity of the air.

Humidity is a vital abiotic climatic factor affecting transpiration and evaporation rates.

3

Match both required instruments in the specified order.

The correct sequence is Secchi disc followed by hygrometer.

This combination accurately pairs the aquatic factor (light penetration) and atmospheric factor (humidity) with their respective instruments.

Key Concept

Measurement of Abiotic Ecological Factors

Question 4488Question →

Four newly documented specimens ( $P$ , $Q$ , $R$ , and $S$ ) were logged with their diagnostic cellular characteristics and proposed scientific names in a taxonomy survey:

- **Specimen $P$ **: Unicellular prokaryote with a peptidoglycan cell wall, recorded as *Escherichia Coli*.
- **Specimen $Q$ **: Acellular nucleoprotein particle consisting of RNA and a protein capsid, classified under Kingdom Monera as *Tobacco mosaic virus*.
- **Specimen $R$ **: Multicellular eukaryotic organism with chitinous cell walls exhibiting extracellular saprophytic digestion, recorded as *Rhizopus stolonifer*.
- **Specimen $S$ **: Non-vascular thalloid plant bearing rhizoids, classified under Division Tracheophyta as *Marchantia polymorpha*.

Which specimen's record strictly adheres to both the Linnaean rules of binomial nomenclature and accurate taxonomic classification?

Specimen R, because *Rhizopus stolonifer* exhibits fungal characteristics and its species epithet is correctly written in lowercase.

Specimen P, because *Escherichia Coli* represents a typical unicellular prokaryote with a peptidoglycan cell wall.

Specimen Q, because *Tobacco mosaic virus* is an acellular entity appropriately categorized within Kingdom Monera.

Specimen S, because *Marchantia polymorpha* is a thalloid organism correctly assigned to Division Tracheophyta.

Show answer & explanation

Answer: Specimen R, because *Rhizopus stolonifer* exhibits fungal characteristics and its species epithet is correctly written in lowercase.

Answer

Specimen R is correctly classified and formatted as *Rhizopus stolonifer*.

The choice identifying Specimen R as correct is accurate because *Rhizopus stolonifer* fulfills all fungal diagnostic traits (chitinous cell wall, saprophytic nutrition) and strictly follows Linnaean binomial rules by capitalizing only the Genus name (*Rhizopus*) while keeping the species epithet (*stolonifer*) in lowercase.

Step-by-Step Solution

1

Evaluate binomial nomenclature formatting rules for each scientific name.

Specimen P (*Escherichia Coli*) violates the rule requiring specific epithets to start with a lowercase letter (*Escherichia coli*).

Under Linnaean binomial nomenclature rules, the genus name begins with an uppercase letter while the species epithet must be entirely in lowercase.

2

Assess the taxonomic group boundary of viruses (Specimen Q).

Specimen Q (*Tobacco mosaic virus*) is acellular and cannot be placed inside Kingdom Monera.

Kingdom Monera contains cellular prokaryotic organisms. Viruses are obligate intracellular non-cellular parasites.

3

Assess the division placement of non-vascular plants (Specimen S).

Specimen S (*Marchantia polymorpha*) is a bryophyte, not a tracheophyte.

Division Tracheophyta is restricted to vascular plants possessing true xylem and phloem, whereas bryophytes lack true vascular bundles.

4

Confirm diagnostic features and binomial rules for Specimen R.

Specimen R (*Rhizopus stolonifer*) possesses chitinous cell walls, saprophytic nutrition (Kingdom Fungi), and a properly capitalized and formatted binomial name.

All structural traits and nomenclature formatting rules are fully satisfied.

Key Concept

Rules of Binomial Nomenclature and Kingdom Diagnostic Features

Estimated Time:1m 30s

Question 4489Question →

When a few drops of aqueous sodium hydroxide are added to a solution of aluminium chloride, a white gelatinous precipitate forms. Upon adding excess sodium hydroxide, the precipitate dissolves to produce a clear, colorless solution. Which chemical species is present in the final clear solution?

Tetrahydroxoaluminate(III) ion,

[Al(OH)_4]^-

Aluminium hydroxide,

Al(OH)_3

Aluminium oxide,

Al_2O_3

Hexaaquaaluminium(III) ion,

[Al(H_2O)_6]^{3+}

Show answer & explanation

Answer: Tetrahydroxoaluminate(III) ion, $[Al(OH)_4]^-$

Answer

The tetrahydroxoaluminate(III) ion,

[Al(OH)_4]^-

Aluminium hydroxide,

Al(OH)_3

, is an amphoteric hydroxide. It initially precipitates as a white gelatinous solid when alkali is added. When excess strong alkali like sodium hydroxide is added, it acts as an acid and reacts further with hydroxide ions to form the soluble complex ion, tetrahydroxoaluminate(III),

[Al(OH)_4]^-

, yielding a clear solution.

Step-by-Step Solution

1

Identify the initial precipitation reaction

Adding a few drops of

OH^-

ions precipitates white gelatinous aluminium hydroxide:

Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)

Aluminium ions react with hydroxide ions to form insoluble aluminium hydroxide.

2

Apply the amphoteric property of aluminium hydroxide with excess strong base

Adding excess sodium hydroxide causes

Al(OH)_3

to act as an acid, dissolving according to:

Al(OH)_3(s) + OH^-(aq) \rightarrow [Al(OH)_4]^-(aq)

Amphoteric hydroxides react with excess alkalis to form soluble complex aluminate ions.

Key Concept

Amphoteric nature of aluminium compounds

Estimated Time:1m 0s

Question 4490Question →

A botanical research team performed a ringing (girdling) experiment on a woody dicotyledonous plant by removing a complete ring of bark down to the vascular cambium layer near the base of the main stem. After several weeks, a distinct swelling was observed immediately above the ring, while the roots gradually died due to sugar starvation. Which transport process and direction were disrupted to cause this specific outcome?

Downward translocation of organic solutes via phloem tissue

Upward translocation of manufactured sugars via xylem vessels

Downward movement of mineral salts and water via xylem vessels

Bidirectional flow of inorganic nutrients via sieve tube elements

Show answer & explanation

Answer: Downward translocation of organic solutes via phloem tissue

Answer

Downward translocation of organic solutes via phloem tissue

Ringing removes phloem, which is responsible for translocating organic food solutes manufactured during photosynthesis downward from leaves to roots. Blocking this path leads to solute accumulation above the ring, causing swelling, and starves roots of carbohydrates.

Step-by-Step Solution

1

Analyze the structural effect of ringing (girdling) a woody stem down to the cambium

Removing the outer layers (bark, cortex, and phloem) removes the phloem tissue while leaving the deeper xylem intact.

Phloem is located towards the periphery of vascular bundles in dicot stems.

2

Identify the primary substance and direction of transport affected by phloem removal

Organic products of photosynthesis (sucrose) synthesized in leaves are transported downward to root sinks via phloem.

Roots act as non-photosynthetic sink organs reliant on shoot sources for metabolic energy.

3

Evaluate the physiological consequences observed in the stem and roots

Accumulated sucrose and auxins above the girdle cause cell proliferation and swelling; root starvation occurs due to lack of carbohydrate supply.

Phloem transport blockade prevents organic food from crossing the ringed zone.

Key Concept

Phloem Translocation and Girdling Experiments

Question 4491Question →

In the binomial system of nomenclature established by Carl Linnaeus, every organism is assigned a two-part scientific name. Which of the following is the correctly written scientific name for the housefly?

*Musca domestica*

*Musca Domestica*

*musca domestica*

Musca domestica

Show answer & explanation

Answer: *Musca domestica*

Answer

The scientific name formatted as *Musca domestica* is correct.

The scientific name starting with a capitalized genus name followed by a lowercase specific epithet in italics complies strictly with the universal rules of binomial nomenclature.

Step-by-Step Solution

1

Identify the basic rules of Linnaean binomial nomenclature.

Binomial nomenclature requires a generic name (Genus) followed by a specific epithet (species).

Scientific names follow universal standards established by international codes of biological nomenclature.

2

Apply capitalization and typography rules to the options.

The genus name (*Musca*) must begin with a capital letter, the species epithet (*domestica*) must be entirely lowercase, and the entire name must be italicized in print.

This standardized formatting prevents confusion across different languages and regions.

Key Concept

Rules of Binomial Nomenclature

Question 4492Question →

A mature dicotyledonous tree undergoes a complete girdling (ringing) experiment in which a continuous ring of outer bark, cortex, and phloem tissue down to the vascular cambium is removed around the lower trunk. Several days after the treatment, carbohydrate accumulation is observed above the ringed area, while root cellular respiration decreases and roots eventually begin to die before the leaves show signs of wilting. Which of the following best explains why root decay precedes foliage wilting in this experiment?

Organic nutrients manufactured in the leaves are translocated downwards via the phloem to sustain root metabolic activities, whereas xylem vessels located inner to the cambium remain intact to continue upward water transport.

Water and inorganic ions are transported unidirectionally downwards through the phloem to the roots, which halts immediately when the outer bark is severed.

Girdling breaks the continuous transpiration pull and cohesion-tension column within phloem sieve tubes required to draw water from the soil.

Phloem vessels transport mineral salts upward to the leaves while xylem translocates manufactured carbohydrates bidirectionally to the root system.

Show answer & explanation

Answer: Organic nutrients manufactured in the leaves are translocated downwards via the phloem to sustain root metabolic activities, whereas xylem vessels located inner to the cambium remain intact to continue upward water transport.

Answer

Organic nutrients manufactured in the leaves are translocated downwards via the phloem to sustain root metabolic activities, whereas xylem vessels located inner to the cambium remain intact to continue upward water transport.

In dicotyledonous stems, phloem is situated peripheral to the vascular cambium while xylem lies deeper toward the center. Girdling interrupts the downward translocation of photosynthetic products (sucrose) in the phloem, depriving root tissues of essential respiratory substrates. Because xylem remains intact inner to the cambium, water and dissolved mineral transport up to the leaves continues temporarily, allowing leaves to stay turgid until root cellular collapse eventually stops water uptake.

Step-by-Step Solution

1

Identify the structural tissues removed during girdling.

Girdling removes the bark, cortex, and phloem layer situated outside the vascular cambium, while leaving the internal xylem undisturbed.

Understanding vascular anatomy is essential to determine which transport pathway is interrupted.

2

Analyze the direction and content of transport in phloem versus xylem.

Phloem conducts sucrose and organic compounds downwards (from leaf source to root sink). Xylem conducts water and mineral salts upward (from roots to leaves).

This establishes which physiological process fails when phloem is cut.

3

Deduce the sequence of physiological impacts on roots versus leaves.

Removing phloem stops downward sucrose supply to roots, leading to starvation of root cells and decay. Leaves continue receiving water through intact xylem for transpiration until roots lose functional integrity.

This explains why root death occurs before leaf wilting.

Key Concept

Differential Vascular Function in Girdling Experiments

Question 4493Question →

A current of $2.0\text{ A}$ is passed through an aqueous solution of copper(II) tetraoxosulfate(VI) for $965\text{ seconds}$ . What is the mass of copper deposited at the cathode? $[F = 96,500\text{ C mol}^{-1},\text{ Cu} = 64]$

0.64\text{ g}

1.28\text{ g}

2.56\text{ g}

0.32\text{ g}

Show answer & explanation

Answer: $0.64\text{ g}$

Answer

The mass of copper deposited at the cathode is

0.64\text{ g}

.

The total charge passed is

Q = 2.0\text{ A} \times 965\text{ s} = 1930\text{ C}

. Dividing by Faraday's constant yields

0.02\text{ mol}

of electrons. Because copper(II) ions require

2\text{ moles}

of electrons per mole of copper metal (

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

), the amount of copper deposited is

0.01\text{ mol}

, which corresponds to

0.01 \times 64 = 0.64\text{ g}

.

Step-by-Step Solution

1

Calculate total quantity of electricity (

Q

) passed

Q = I \times t = 2.0\text{ A} \times 965\text{ s} = 1930\text{ C}

Faraday's first law relates charge to current and time.

2

Calculate the moles of electrons transferred

\text{Moles of } e^- = \frac{Q}{F} = \frac{1930\text{ C}}{96,500\text{ C mol}^{-1}} = 0.02\text{ mol } e^-

One mole of electrons carries one Faraday (

96,500\text{ C}

) of charge.

3

Determine moles of copper deposited using electrode reduction half-equation

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \implies n(\text{Cu}) = \frac{0.02\text{ mol } e^-}{2} = 0.01\text{ mol}

Reduction of one

\text{Cu}^{2+}

ion requires

2

moles of electrons per mole of copper deposited.

4

Calculate mass of copper deposited

\text{Mass} = n \times \text{Molar Mass} = 0.01\text{ mol} \times 64\text{ g mol}^{-1} = 0.64\text{ g}

Multiplying moles of substance by its molar mass yields mass.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Stoichiometry of Charge

Estimated Time:1m 0s

Question 4494Question →

A complete ring of bark including the phloem tissue is carefully removed from the trunk of a woody dicotyledonous plant while keeping the xylem intact. After several weeks, a distinct swelling is observed in the bark region immediately above the cut ring. Which of the following processes explains the formation of this swelling?

The downward translocation of manufactured organic food is blocked, causing solutes to accumulate above the ring.

The upward transport of water and mineral salts is obstructed, forcing fluids to collect in the stem tissue above the ring.

The bryophyte vascular bundles are severed, preventing the redistribution of inorganic minerals to lower tissues.

Oxygen gas produced during dark photosynthetic reactions is trapped and inflates the outer cortex above the girdle.

Show answer & explanation

Answer: The downward translocation of manufactured organic food is blocked, causing solutes to accumulate above the ring.

Answer

The swelling above the ringed stem is caused by the accumulation of organic food substances translocated downward through the phloem from the leaves.

In girdling (ringing) experiments, removing the phloem severs the transport route for organic compounds (mainly sucrose) synthesized during photosynthesis in leaves. Because xylem remains intact, leaves continue to receive water and produce photosynthates. As manufactured food moves down the stem via phloem, it meets the blocked ring and accumulates immediately above it, causing localized tissue swelling due to organic accumulation and cell division.

Step-by-Step Solution

1

Identify the tissues present in the removed ring of bark.

Removing bark removes the phloem (sieve tubes and companion cells) but leaves the inner xylem vessels intact.

Phloem lies in the outer vascular region of dicot stems, while xylem lies towards the inner core.

2

Determine the transport function and direction of phloem versus xylem.

Phloem translocates manufactured organic food (sucrose, amino acids) bidirectionally/downward from leaves to sink tissues, whereas xylem conducts water and mineral salts upward from roots.

Leaves photosynthesize organic nutrients that must travel downward to feed stem and root cells.

3

Analyze the physical effect of severing phloem tissue.

Organic solutes flowing down from the canopy are blocked at the upper boundary of the cut ring, accumulating and swelling the tissue above the girdle due to localized cell growth and osmotic water intake.

The intact xylem continues supplying water to the leaves, maintaining photosynthesis and translocation down to the point of interruption.

Key Concept

Phloem translocation of organic solutes in plants

Question 4495Question →

Consider the reversible industrial synthesis of ammonia gas: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H = -92\text{ kJ mol}^{-1}$ . A chemical engineer introduces a finely divided iron catalyst into the reaction vessel while maintaining constant temperature and pressure. Which of the following best describes the effect of adding the catalyst on the system?

It selectively lowers the activation energy of the forward reaction, shifting the equilibrium position to the right and increasing the yield of

NH_3

.

It increases the average kinetic energy of the reactant molecules, causing an increase in the equilibrium constant (

K_c

) and a higher yield of

NH_3

.

It lowers the activation energy for both the forward and reverse reactions equally, increasing their rates and reducing the time taken to reach equilibrium without altering the final yield of

NH_3

.

It shifts the position of equilibrium to the left because the forward reaction is exothermic, thereby favoring the decomposition of

NH_3

.

Show answer & explanation

Answer: It lowers the activation energy for both the forward and reverse reactions equally, increasing their rates and reducing the time taken to reach equilibrium without altering the final yield of $NH_3$ .

Answer

Adding a catalyst lowers the activation energy for both forward and reverse reactions by equal amounts, increasing their rates and reducing the time needed to reach dynamic equilibrium without altering the equilibrium yield of ammonia.

A catalyst functions by offering an alternative pathway with a lower activation energy barrier. In a reversible system, this reduction in activation energy applies equally to both the forward and reverse directions. Consequently, the rates of both reactions increase by the same proportion, enabling the system to reach dynamic equilibrium faster without changing the equilibrium composition or the yield of products.

Step-by-Step Solution

1

Analyze the function of a catalyst in chemical kinetics.

A catalyst provides an alternative reaction pathway with a lower activation energy (

E_a

).

Lowering

E_a

allows a greater fraction of reactant molecules to possess sufficient energy to undergo effective collisions per unit time.

2

Evaluate the symmetry of activation energy reduction in reversible reactions.

The catalyst lowers the activation energy barrier for both the forward reaction and the reverse reaction by the exact same amount (

Δ E_a

).

Since the initial reactants and final products remain in the same energy states, the difference between the forward and reverse activation energy barriers (

Δ H

) is unchanged.

3

Determine the impact of the catalyst on chemical equilibrium and product yield.

The rates of both forward and reverse reactions increase by the same factor, so the equilibrium position and equilibrium constant (

K_c

) remain unchanged; only the time to reach equilibrium decreases.

Equal acceleration of both directions leaves the relative equilibrium concentrations of products and reactants unaffected.

Key Concept

Effect of Catalysts on Reaction Kinetics and Chemical Equilibrium

Estimated Time:1m 30s

Question 4496Question →

Consider a diploid organism with the genotype $AaBb$ , where genes $A$ and $B$ are located on separate homologous chromosome pairs. During meiotic prophase I, crossing over does not occur between these loci. Which of the following statements correctly accounts for the relationship between gene loci, alleles, and sister chromatids during gamete formation in this organism?

The alleles

A

and

a

occupy identical loci on homologous chromosomes, whereas identical copies of allele

A

exist on sister chromatids prior to anaphase I segregation.

The alleles

A

and

a

represent alternative structural forms of the same gene located on sister chromatids of a single replicated chromosome.

The phenotype expressed by the heterozygous state

Aa

is always an exact intermediate blending of the homozygous dominant

AA

and homozygous recessive

aa

traits.

The alleles

A

and

B

are codominant alternatives located at the same physical gene locus on a single chromosome pair.

Show answer & explanation

Answer: The alleles $A$ and $a$ occupy identical loci on homologous chromosomes, whereas identical copies of allele $A$ exist on sister chromatids prior to anaphase I segregation.

Answer

The alleles

A

and

a

occupy identical loci on homologous chromosomes, whereas identical copies of allele

A

exist on sister chromatids prior to anaphase I segregation.

Homologous chromosomes contain corresponding gene loci where alternative alleles (

A

and

a

) reside. Following S-phase replication prior to meiosis I, each chromosome consists of two sister chromatids carrying identical copies of its allele (e.g., two copies of

A

on one chromosome and two copies of

a

on its homologue).

Step-by-Step Solution

1

Analyze the relationship between homologous chromosomes and gene loci.

In diploid organisms, homologous chromosomes occur in pairs, carrying genes for the same traits at identical physical locations termed loci. Therefore, alternative forms of a gene (alleles

A

and

a

) are located at corresponding loci on paternal and maternal homologous chromosomes.

Establishing the physical location of alleles on homologous chromosomes is essential to distinguish homologous pairs from replicated chromatids.

2

Analyze the composition of sister chromatids following S-phase replication.

During the S phase of interphase, each chromosome replicates to form two sister chromatids joined at the centromere. Sister chromatids carry genetically identical nucleotide sequences and thus identical alleles (e.g.,

A

and

A

).

Differentiating sister chromatids (identical duplicated strands) from homologous chromosomes (maternal and paternal counterparts) prevents confusing allele variation with chromatid duplication.

3

Evaluate independent assortment and locus position for unlinked genes

A

and

B

.

Genes

A

and

B

reside on distinct chromosome pairs. They represent separate loci rather than alleles of the same gene locus.

Alleles compete for the same locus, whereas separate genes reside at different loci across the genome.

Key Concept

Basic Genetics Terminology and Concepts

Question 4497Question →

An investigation of four organisms collected during a field survey yielded the following observations and proposed names:

- Organism 1: A multicellular saprophyte with chitinous cell walls, named *rhizopus Stolonifer*.
- Organism 2: A unicellular prokaryote with a peptidoglycan cell wall, named *Escherichia coli*.
- Organism 3: An acellular nucleoprotein entity classified under Kingdom Monera as *tobacco Mosaic Virus*.
- Organism 4: A non-flowering plant possessing true xylem and phloem, classified as a bryophyte named *Funaria hygrometrica*.

Based on the principles of biological classification and binomial nomenclature, which organism is correctly described and has its scientific name properly formatted?

Organism 2

Organism 1

Organism 3

Organism 4

Show answer & explanation

Answer: Organism 2

Answer

Organism 2 is correctly described and has its scientific name properly formatted.

Organism 2 is a unicellular prokaryote with a peptidoglycan cell wall, accurately defining a bacterium, and its scientific name *Escherichia coli* is formatted correctly according to Linnaean rules (capitalized genus, lowercase species epithet, italicized).

Step-by-Step Solution

1

Evaluate the binomial nomenclature formatting for each organism.

Organism 1 (*rhizopus Stolonifer*) fails binomial rules because the genus name must be capitalized and the species epithet must be lowercase. Organism 2 (*Escherichia coli*) adheres strictly to the rule.

Binomial nomenclature requires the genus to start with an uppercase letter and the species epithet with a lowercase letter.

2

Assess the cellular classification of acellular entities.

Organism 3 (a virus) cannot belong to Kingdom Monera.

Viruses are acellular nucleoprotein complexes lacking cytoplasm and cell membranes, excluding them from cellular kingdoms.

3

Analyze anatomical characteristics against kingdom and phylum definitions.

Organism 4 (*Funaria hygrometrica*) is a bryophyte, which inherently lacks vascular tissue (xylem and phloem).

True vascular tissues appear in tracheophytes (pteridophytes and spermatophytes), not bryophytes.

4

Synthesize findings to select the fully valid choice.

Only Organism 2 satisfies both correct biological characterization and standard Linnaean naming rules.

Bacteria belong to Monera, possess peptidoglycan cell walls, and *Escherichia coli* is correctly formatted.

Key Concept

Rules of Binomial Nomenclature and Diagnostic Kingdom Traits

Estimated Time:2m 0s

Question 4498Question →

For a reversible gas-phase reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ , the total enthalpy of the reactants is $+120\text{ kJ mol}^{-1}$ . The reaction is exothermic with a standard enthalpy change ( $\Delta H$ ) of $-45\text{ kJ mol}^{-1}$ . In the absence of a catalyst, the activation energy for the reverse reaction ( $E_{a,\text{rev}}$ ) is $+185\text{ kJ mol}^{-1}$ . If a catalyst is introduced that lowers the activation energy barrier by $30\text{ kJ mol}^{-1}$ , what is the activation energy for the catalyzed forward reaction?

+110\text{ kJ mol}^{-1}

+140\text{ kJ mol}^{-1}

+155\text{ kJ mol}^{-1}

+200\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $+110\text{ kJ mol}^{-1}$

Answer

+110\text{ kJ mol}^{-1}

The correct answer is

+110\text{ kJ mol}^{-1}

. For an exothermic reaction with

\Delta H = -45\text{ kJ mol}^{-1}

and reverse activation energy

E_{a,\text{rev}} = +185\text{ kJ mol}^{-1}

, the uncatalyzed forward activation energy is

E_{a,\text{fwd}} = E_{a,\text{rev}} + \Delta H = 185 - 45 = +140\text{ kJ mol}^{-1}

. Adding a catalyst lowers this activation energy barrier by

30\text{ kJ mol}^{-1}

, yielding

140 - 30 = +110\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Calculate the uncatalyzed forward activation energy (

E_{a,\text{fwd, uncatalyzed}}

)

E_{a,\text{fwd, uncatalyzed}} = E_{a,\text{rev}} + \Delta H = +185\text{ kJ mol}^{-1} + (-45\text{ kJ mol}^{-1}) = +140\text{ kJ mol}^{-1}

For any chemical system, the relationship between forward activation energy, reverse activation energy, and enthalpy change is

\Delta H = E_{a,\text{fwd}} - E_{a,\text{rev}}

.

2

Apply the catalyst reduction to the forward activation energy

E_{a,\text{fwd, catalyzed}} = E_{a,\text{fwd, uncatalyzed}} - 30\text{ kJ mol}^{-1} = 140 - 30 = +110\text{ kJ mol}^{-1}

A catalyst lowers both forward and reverse activation energies by equal amounts, reducing the energy barrier height.

Key Concept

Relationship between forward activation energy, reverse activation energy, enthalpy change, and catalyst effect in energy profile diagrams

Alternative Method

Calculate the energy levels directly: Reactants =

+120\text{ kJ mol}^{-1}

. Products =

120 + (-45) = +75\text{ kJ mol}^{-1}

. Uncatalyzed Transition State = Products +

E_{a,\text{rev}} = 75 + 185 = +260\text{ kJ mol}^{-1}

. Catalyzed Transition State =

260 - 30 = +230\text{ kJ mol}^{-1}

. Catalyzed

E_{a,\text{fwd}} = \text{Catalyzed Transition State} - \text{Reactants} = 230 - 120 = +110\text{ kJ mol}^{-1}

.

Estimated Time:2m 0s

Question 4499Question →

The standard reduction potentials for iron and copper half-cells are given as follows:

\text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.44\text{ V}

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\text{ V}

What is the standard electromotive force ( $E^\circ_{\text{cell}}$ ) for the overall reaction $\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s)$ , and is the reaction spontaneous under standard conditions?

+0.78\text{ V}

, spontaneous

-0.78\text{ V}

, non-spontaneous

-0.10\text{ V}

, non-spontaneous

+0.10\text{ V}

, spontaneous

Show answer & explanation

Answer: $+0.78\text{ V}$ , spontaneous

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) is

+0.78\text{ V}

, and the reaction is spontaneous.

In the given reaction,

\text{Cu}^{2+}

ions are reduced to copper metal at the cathode (

E^\circ = +0.34\text{ V}

), while

\text{Fe}

metal is oxidized to

\text{Fe}^{2+}

ions at the anode (

E^\circ = -0.44\text{ V}

). Using the standard formula

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

, we calculate

E^\circ_{\text{cell}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}

. Because the cell potential is positive, the reaction is spontaneous under standard conditions.

Step-by-Step Solution

1

Identify the reduction and oxidation half-reactions

Cathode (reduction):

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

with

E^\circ_{\text{cathode}} = +0.34\text{ V}

. Anode (oxidation):

\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-

with

E^\circ_{\text{anode}} = -0.44\text{ V}

.

Copper ions gain electrons (reduction at cathode) while iron metal loses electrons (oxidation at anode).

2

Calculate the standard cell potential (

E^\circ_{\text{cell}}

)

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}

.

The cell potential is the difference between the reduction potential of the cathode species and that of the anode species.

3

Determine reaction spontaneity

Since

E^\circ_{\text{cell}} = +0.78\text{ V} > 0

, the reaction is spontaneous.

A positive standard cell potential indicates a thermodynamically feasible (spontaneous) redox reaction under standard conditions.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 15s

Question 4500Question →

A biology student wishes to measure the moisture content of a soil sample collected from a farmland. Arrange the following laboratory procedure steps in the correct sequential order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of steps is: (1) Weigh the freshly collected soil sample immediately to determine its initial fresh mass, (2) Place the soil sample in an oven maintained at

105^\circ\text{C}

to evaporate moisture, (3) Cool the sample inside a desiccator and re-weigh it until a constant dry mass is obtained, and (4) Calculate the difference between the fresh mass and constant dry mass to express water content as a percentage.

Measuring soil moisture requires establishing an initial fresh mass, evaporating the water by oven drying at

105^\circ\text{C}

, cooling in a moisture-free desiccator until a constant dry mass is reached, and finally computing the percentage loss of mass.

Step-by-Step Solution

1

Measure baseline fresh mass

Obtain initial mass containing both dry soil solids and water.

A starting mass is required to compute the total mass lost as water.

2

Evaporate water content

Water leaves the sample as steam.

Oven drying at

105^\circ\text{C}

removes water without decomposing organic components.

3

Cool safely and verify complete drying

Obtain true dry mass.

Desiccators prevent re-absorption of atmospheric moisture; achieving constant mass ensures all water was removed.

4

Calculate moisture percentage

Determine soil moisture percentage using

\frac{\text{Fresh Mass} - \text{Dry Mass}}{\text{Fresh Mass}} \times 100\%

.

Quantifies the edaphic factor (water content) relative to the fresh sample mass.

Key Concept

Measurement of Edaphic Factors (Soil Water Content)

Estimated Time:1m 0s

Answer

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